How do you integrate #1/(9x^2+4)^2 dx#?

Question

Christopher Agresta · Accepted Answer

# 1/48{arc tan((3x)/2)+(6x)/(9x^2+4)}+C#.

Let, #I=int1/(9x^2+4)^2dx#. #9x^2+4=(3x)^2+2^2#, suggests that, #3x=2tany# substitution may work. So, let, #3x=2tany. :. 3dx=2sec^2ydy#. #:. I=int1/(4tan^2y+4)^2*2/3sec^2ydy#, #=2/48intsec^2y/sec^4ydy#, #=1/24intcos^2ydy#, #=1/24int(1+cos2y)/2dy#, #=1/48{y+(sin2y)/2}#, #=1/48{y+1/2*(2tany)/(1+tan^2y)}#, Here, #3x=2tany rArr y=arc tan((3x)/2)#. #:. I=1/48{arc tan((3x)/2)+((3x)/2)/(1+((3x)/2)^2)}#. # rArr I=1/48{arc tan((3x)/2)+(6x)/(9x^2+4)}+C#.

Evelyn Agard · Answer

#int \frac{1}{(9 x^2 + 4)^2} dx = \frac{1}{48} [\frac{6 x}{9 x^2 + 4} + tan^{-1} \frac{3 x}{2} ] + C#

As a first simplification, let's take the 9 out of the denominator:

int \frac{1}{(9 x^2 + 4)^2} dx = \frac{1}{81} int \frac{1}{(x^2 + \frac{4}{9})^2} dx #.

If we have a look at the denominator now, then it somehow reminds us of the (RHS of the) trigonometric identity #sec^2 x = tan^2 x + 1#, and besides this, we remember that #(tan x)' = sec^2 x#. Therefore, we make the substitution #x \mapsto g ( t )# with

#g ( t ) = \frac{2}{3} tan t#, #g' ( t ) = \frac{2}{3} sec^2 t#, and #g^{-1} ( x ) = tan^{-1} \frac{3 x}{2}#.

With this, we get

[ \frac{1}{81} int \frac{1}{(\frac{4 tan^2 t}{9} + \frac{4}{9})^2} \cdot \frac{2}{3} sec^2 t\ dt ]_{t = tan^{-1} \frac{3 x}{2}} =

= [ \frac{1}{24} int \frac{1}{(tan^2 t + 1)^2} \cdot sec^2 t\ dt ]_{t = tan^{-1} \frac{3 x}{2}} #.

Using the trigonometric identity mentioned before, this is the same as

[ \frac{1}{24} int \frac{1}{sec^4 t} \cdot sec^2 t\ dt ]_{t = tan^{-1} \frac{3 x}{2}} =

= [ \frac{1}{24} int cos^2 t\ dt ]_{t = tan^{-1} \frac{3 x}{2}} #.

According to the double-angle formula for the cosine, #cos ( 2t ) = 2 cos^2 t - 1#, this is the same as

[ \frac{1}{48} int cos( 2 t) + 1\ dt ]_{t = tan^{-1} \frac{3 x}{2}} #,

which is easily evaluated to be

\frac{1}{48} [ \frac{sin(2 t)}{2} + t ]_{t = tan^{-1} \frac{3 x}{2}} + C #.

Lastly, we use the double-angle formula for the sine,

#sin (2t) = (2 tan t) / (1 + tan^2 t)#,

and thus arrive at the result

\frac{1}{48} [ \frac{tan t}{1 + tan^2 t} + t ]_{t = tan^{-1} \frac{3 x}{2}} + C =

= \frac{1}{48} [ \frac{\frac{3 x}{2}}{1 + \frac{9 x^2}{4}} + tan^{-1} \frac{3 x}{2} ] + C =

= \frac{1}{48} [ \frac{6 x}{9 x^2 + 4} + tan^{-1} \frac{3 x}{2} ] + C #.

Christopher Agresta · Answer

undefined To integrate $ \frac{1}{{(9x^2 + 4)}^2} \, dx $, you can use a trigonometric substitution. Let $ x = \frac{2}{3} \tan(\theta) $. Then, $ dx = \frac{2}{3} \sec^2(\theta) \, d\theta $. Substitute these into the integral and simplify to express the integrand in terms of $ \theta $. After simplification, the integral should become a standard trigonometric integral which can be evaluated using trigonometric identities and integration techniques. Finally, resubstitute $ x $ in terms of $ \theta $ to get the final result.